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HW 6 Calc

HW 6 Calc - garcia(jjg2564 – HW 06 – gualdani –(56410...

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Unformatted text preview: garcia (jjg2564) – HW 06 – gualdani – (56410) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Find the x-intercept of the tangent line to the graph of f ( x ) = sin x + 3 cos x at the point (0 , f (0)). 1. x-intercept = 3 4 2. x-intercept =- 3 correct 3. x-intercept = 3 4. x-intercept = 1 3 5. x-intercept =- 1 4 6. x-intercept =- 1 3 Explanation: When f ( x ) = sin x + 3 cos x , then f ′ ( x ) = cos x- 3 sin x . Thus at x = 0, f (0) = 3 , f ′ (0) = 1 . So by the Point-Slope formula, an equation for the tangent line at (0 , f (0)) is y- 3 = 1( x- 0) , which after rearranging become y = x + 3 . Consequently, the tangent line to the graph of f at (0 , f (0)) has x-intercept =- 3 . keywords: tangent, trig function, sin, cos, trig derivative, intercept, point-slope formula, 002 10.0 points Find the derivative of f when f ( x ) = sin x 2 + 3 cos x . 1. f ′ ( x ) = 2 cos x- 3 2 + 3 cos x 2. f ′ ( x ) = 2 cos x- 3 (2 + 3 cos x ) 2 3. f ′ ( x ) = 2 cos x + 3 (2 + 3 cos x ) 2 correct 4. f ′ ( x ) = 2 cos x + 3 2 + 3 cos x 5. f ′ ( x ) = 2 cos x + 3 sin 2 x (2 + 3 cos x ) 2 Explanation: By the Quotient Rule, f ′ ( x ) = cos x (2 + 3 cos x ) + 3 sin x sin x (2 + 3 cos x ) 2 = 2 cos x + 3(cos 2 x + sin 2 x ) (2 + 3 cos x ) 2 . But cos 2 x + sin 2 x = 1, so f ′ ( x ) = 2 cos x + 3 (2 + 3 cos x ) 2 . 003 10.0 points Determine the derivative of f ( x ) = tan x 5 + sec x . 1. f ′ ( x ) = tan x (5 sec x + 1) 5 + sec x garcia (jjg2564) – HW 06 – gualdani – (56410) 2 2. f ′ ( x ) = sec x (5 sec x- 1) (5 + sec x ) 2 3. f ′ ( x ) = 5 sec x tan x 5 + sec x 4. f ′ ( x ) = tan x (5 sec x- 1) 5 + sec x 5. f ′ ( x ) = sec x (5 sec x + 1) (5 + sec x ) 2 correct 6. f ′ ( x ) = 5 sec 2 x (5 + sec x ) 2 Explanation: Since (tan x ) ′ = sec 2 x, (sec x ) ′ = sec x tan x , the Quotient Rule ensures that f ′ ( x ) = sec 2 x (5 + sec x )- sec x tan 2 x (5 + sec x ) 2 = 5 sec 2 x + sec x (sec 2 x- tan 2 x ) (5 + sec x ) 2 . But sec 2 x = 1 + tan 2 x . Consequently, f ′ ( x ) = sec x (5 sec x + 1) (5 + sec x ) 2 . 004 10.0 points Find an equation for the tangent line to the graph of f at the point P ( π 4 , f ( π 4 )) when f ( x ) = 5 tan x . 1. y = 15 x + 4 parenleftBig 1- π 2 parenrightBig 2. y = 10 x + 5 parenleftBig 1- π 2 parenrightBig correct 3. y = 2 x + 10 parenleftBig 1- π 4 parenrightBig 4. y = 18 x + 13 parenleftBig 1- π 2 parenrightBig 5. y = 9 x + 12 parenleftBig 1- π 4 parenrightBig Explanation: When x = π 4 , then f ( x ) = 5, so P = ( π 4 , 5)....
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HW 6 Calc - garcia(jjg2564 – HW 06 – gualdani –(56410...

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